Chem 211-10
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Experiment 7
Part A
How Can We
Determine the
Structure of an
Organic Compound?
A. Pre-lab Preparation Assignment:
1. Read: This handout
2.
Read: Lab Manual Part V. Appendix D1.
Padías pp. 98-102
3. Read the Mass Spectrometry Background and Introduction to Mass Spectrometry links on the
Experiment 7A page of the course website.
4. Work Through: The Introduction to Mass Spectrometry and Structural Analysis activity
below and be ready to respond to the questions at the bottom of the last page.
5. Complete the electronic Prelab Questions (course website) by 8:00 PM, Monday, November
1 on the Course Blackboard Site as explained on the Course Information page.
6. Consider possible claims and warrants for the QOW (See below)
7. In your lab notebook: (See Lab Manual pp. 15-17 for format.)
a. Enter the title of the experiment, name of group members and QOW on the appropriate
first left-hand page.
b. Up-date the Table of Contents.
7. Bring your notebook and your initial claims and warrants concerning the QOW to the AM
lab discussion period on Tuesday, November 2.
B. Objectives:
1. To gain experience with collecting mass spectra of solid compounds.
2. To gain expertise with analyzing IR and mass spectral data to determine the structure of
organic compounds.
C. Question of the Day:
How Do Organic Compounds Interact with High Energy Electrons
and What Structural Information Can We Obtain from the Results?
D. Key Terms/concepts/techniques:
 Mass spectrometry
 Base Peak
 Molecular ion
Experiment 7A
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211-10
E. Introduction to Mass Spectrometry and Structural Analysis
1. The Mass Selective Detector (MSD).
When a neutral substance enters that mass spectrometer, it is bombarded with high energy
electrons, which cause the substance to ionize as illustrated in Equation 1.
[M] +. + 2 ee + M
Equation 1: Electron Impact Ionization
The species M is both a cation (has a positive charge) and a free radical (has an unpaired
electron) and is referred to as a radical cation. The mass of this first formed radical cation is
equal to that of the original molecule entering the mass spectrometer and is referred to as the
molecular ion.
The molecular ion, because it retains considerable energy from the initial collision, will tend
to break down (fragment) into smaller fragments. These fragmentations usually result from
breakage of one bond dividing the original structure into two pieces one having a positive
charge (a cation) and the other containing the unpaired electron (a free radical). The mass
spectrometer detects only the positive ions and not the free radicals.
2. Characteristics of Mass Spectra.
For our purposes in this activity it is important to know the following:
a. The mass spectrometer detects only positive ions and separates them according to their
mass to charge ratio, m/z.
b. Most ions produced on ionization in the electron beam are unipositive so numerically,
m/z = mass of the ion.
c. The mass of the molecular ion can usually be taken as the molecular weight of the
substance.
d. The mass spectrometer can separate ions that have different isotopes of their atoms.
e. The most intense peak in the spectrum is called the base peak and represents the ion of
highest abundance and therefore the highest intensity.
f. Knowledge of the isotope composition of the more common elements is essential to the
interpretation of mass spectral data. See Figure 1
Figure 1: Table of Isotope Composition of Some Common Elements
Experiment 7A
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3. Initial Analysis of Molecular Ions.
a. Detection of Halogens in Compounds from Molecular Ions:
• The presence of bromine in a compound leads to two molecular ions (with 35Br79 or
with 35Br81) differing by 2 mass units. For a molecule with one bromine atom, what
should be the relative intensities of these two molecular ions?
• The presence of chlorine in a compound leads to two molecular ions (with 17Cl35 or
with 17Cl37) differing by 2 mass units. For a molecule with one chlorine atom, what
should be the relative intensities of these two molecular ions?
b. Additional Molecular Ion Considerations.
As we learned in developing our methods for calculating double bond equivalents:
• The general formula for a hydrocarbon is CnH2n+2. Considering the mass of carbon
and hydrogen, predict if the molecular ion of a hydrocarbon should be an even or
odd number. (choose one) Provide a brief warrant supporting your claim.
• A molecule with one oxygen atom adds a total mass of 16 to the mass of a
corresponding the hydrocarbon. That makes the molecular ion of a compound
containing an O an (even or odd) number (choose one). Provide a brief warrant
supporting your claim.
• A molecule with one nitrogen atom also has an additional hydrogen atom compared
to a corresponding the hydrocarbon, a total mass of 15 (14 + 1). That makes the
molecular ion of a compound containing an N an (even or odd) number (choose
one). Provide a brief warrant supporting your claim.
• A bromine atom replaces a hydrogen atom in a hydrocarbon. That makes the two
molecular ions of a compound containing one Br:
(1) Both even numbers.
(2) Both odd numbers.
(3) One odd number and one even number
• A chlorine atom replaces a hydrogen atom in a hydrocarbon. That makes the two
molecular ions of a compound containing one Cl:
(1) Both even numbers.
(2) Both odd numbers.
(3) One odd number and one even number